The problem statement is true. Here is the proof, Let's call the triangle without exactly $10$ red points inside 'bad triangle'. For any $4$ blue points, if there is a bad triangle among the triangles formed by these points, let's call these $4$ points the 'Fatih Terim group'. Claim: At least $2$ of the $4$ triangles formed by a Fatih Terim group, are bad triangles. Proof: Assume the contrary. Let there be points $A, B, C, D,$ exactly 1 of the triangles formed by them is a bad triangle. WLOG say it's $\triangle ABC$. Then $\triangle ABD, \triangle ACD$ and $\triangle BCD$ must contain $10$ red points. Which implies $\triangle ABC$ contains $10$ red points too ( simply $\triangle BCD +\triangle ABD -\triangle ADC$). So our assumption was wrong. Now assume there exist a bad triangle $ABC$. By our claim, for any $4$ points to be formed with each of the remaining 17 points and $A,B,C$ will have at least one more bad triangle. Obviously these triangles cannot be same. So there must be at least $17+1=18$ bad triangle. But the total number of triangles formed by blue points is $\binom{20}{3}=1140 < 1123+18$ impossible. So our assumption was wrong. $\square$